Question

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are $14$ cm, $10$ cm and $6$ cm. Find the area of the triangle.

• A. $300\sqrt{2}c{m}^2$
  
• B. $300\sqrt{3}c{m}^2$
• C. $600\sqrt{3}c{m}^2$
• D. $600\sqrt{2}c{m}^2$

Answer 1

The correct option is B $300\sqrt{3}c{m}^2$

Let each side of $㎝$ equilateral triangle ABC be ${}^{\prime }{a}^{\prime }㎝$

Now, $ar△OAB=\frac{1}{2}\times AB\times OP$$=\frac{1}{2}\times a\times 14=7a㎠\to 1$

$ar△OBC=\frac{}{}\times BC\times OQ$

$=\frac{1}{2}\times a\times 10=5a㎠\to 2$

$ar△OAC=\frac{1}{2}\times AC\times OR=\frac{1}{2}\times a\times 6=3a㎠\to 3$

$\therefore ar△ABC=1+2+3=7a+5a+3a=15a㎠$

Also area of equilateral triangle $ABC=\frac{\sqrt{3}}{4}{a}^2$

Now, $\frac{\sqrt{3}}{4}{a}^2=15a\Rightarrow a=\frac{15\times 4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{60\sqrt{3}}{3}=20\sqrt{3}㎝$

Now, $ar△ABC=\frac{\sqrt{3}}{4}\times {\left(20\sqrt{3}\right)}^2=300\sqrt{3}㎠$